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Hello,
I hope somebody can help me with this question.
I have to prove that time dilation is a reciprocal effect by only using the Postulates of Minimal Relativity Theory:
1. Let I be an inertial frame, then I' moving with a constant velocity in a rectilinear motion with respect to I will also be an inertial frame.
2. The transformation from I to I' is continuous in v.
3. The motion is reciprocal. If I' moves with v wrt to I then I moves with -v wrt to I'
2. Homework Equations
I formulated transformations from I to I' through the following equations:
[itex]t'=T(x,v,t)[/itex]
[itex]x'=X(x,v,t)[/itex]
[itex]t=T(x',-v,t')[/itex]
[itex]x=X(x',-v,t')[/itex]
Where x,x',v,v' are vector quantities.
Now here are a few questions I have regarding some assumptions we can make about the transformations T and X:
I read that one of the conditions for an inertial frame is that it describes time and space homogeneously and isotropically. My question is whether the transformation T and X will then have to be linear mappings in x and t to conserve homogeneity and isotropy? Is there non-linear mapping which fulfils these criterions?
So basically i used the relativity principle and differentiated the x and x' equation wrt to t and t' respectively.
Where x is the position of frame I wrt. I' and x' the position of frame I' wrt I.
[itex]\frac{dx'}{dt'}=\left(v\nabla X(x,v,t)+\frac{\partial X(x,v,t)}{\partial t}\right)\frac{dt}{dt'}[/itex]
[itex]\frac{dx}{dt}=\left(-v\nabla ' X(x',-v,t')+\frac{\partial X(x',-v,t')}{\partial t'}\right)\frac{dt'}{dt}[/itex]
By the principle of relativity we know that:
[itex]\frac{dx'}{dt'}=-\frac{dx}{dt}[/itex]
And I bascially want to prove:
[itex]\frac{dt}{dt'}=\frac{dt'}{dt}[/itex]
Unity might be a solution but there is another solution which seems paradoxical, but is actually not if you keep in mind that t and t' each mean two different things on both sides of the equation.
If I don't rotate I' wrt to I then [itex]\nabla X(x,v,t)=\nabla ' X(x',-v,t')[/itex]
My guess is that if i rotated it then I would also have to rotate v in the other direction and the rotation matrices would cancel. Then somehow I need to show that:
[itex]\frac{\partial X(x,v,t)}{\partial t}=-\frac{\partial X(x',-v,t')}{\partial t'}[/itex]
From the Lorentz transformation I know that this is true. since the only term which contains a t in the x' transformation is the vt term. But I don't know whether it is a valid argument to say that X will have to be a linear combination of terms of which one is vt.
I hope somebody can help me with this question.
Homework Statement
I have to prove that time dilation is a reciprocal effect by only using the Postulates of Minimal Relativity Theory:
1. Let I be an inertial frame, then I' moving with a constant velocity in a rectilinear motion with respect to I will also be an inertial frame.
2. The transformation from I to I' is continuous in v.
3. The motion is reciprocal. If I' moves with v wrt to I then I moves with -v wrt to I'
2. Homework Equations
I formulated transformations from I to I' through the following equations:
[itex]t'=T(x,v,t)[/itex]
[itex]x'=X(x,v,t)[/itex]
[itex]t=T(x',-v,t')[/itex]
[itex]x=X(x',-v,t')[/itex]
Where x,x',v,v' are vector quantities.
Now here are a few questions I have regarding some assumptions we can make about the transformations T and X:
I read that one of the conditions for an inertial frame is that it describes time and space homogeneously and isotropically. My question is whether the transformation T and X will then have to be linear mappings in x and t to conserve homogeneity and isotropy? Is there non-linear mapping which fulfils these criterions?
The Attempt at a Solution
So basically i used the relativity principle and differentiated the x and x' equation wrt to t and t' respectively.
Where x is the position of frame I wrt. I' and x' the position of frame I' wrt I.
[itex]\frac{dx'}{dt'}=\left(v\nabla X(x,v,t)+\frac{\partial X(x,v,t)}{\partial t}\right)\frac{dt}{dt'}[/itex]
[itex]\frac{dx}{dt}=\left(-v\nabla ' X(x',-v,t')+\frac{\partial X(x',-v,t')}{\partial t'}\right)\frac{dt'}{dt}[/itex]
By the principle of relativity we know that:
[itex]\frac{dx'}{dt'}=-\frac{dx}{dt}[/itex]
And I bascially want to prove:
[itex]\frac{dt}{dt'}=\frac{dt'}{dt}[/itex]
Unity might be a solution but there is another solution which seems paradoxical, but is actually not if you keep in mind that t and t' each mean two different things on both sides of the equation.
If I don't rotate I' wrt to I then [itex]\nabla X(x,v,t)=\nabla ' X(x',-v,t')[/itex]
My guess is that if i rotated it then I would also have to rotate v in the other direction and the rotation matrices would cancel. Then somehow I need to show that:
[itex]\frac{\partial X(x,v,t)}{\partial t}=-\frac{\partial X(x',-v,t')}{\partial t'}[/itex]
From the Lorentz transformation I know that this is true. since the only term which contains a t in the x' transformation is the vt term. But I don't know whether it is a valid argument to say that X will have to be a linear combination of terms of which one is vt.